Latest Edit
Great thanks to @Quanto for giving us the complete solution to ALL even powers using two wonderful factorizations
\begin{align}
&1+x^{4m}= \prod_{k=1}^m \left(1+2x^2\cos\frac{(2k-1)\pi}{2m}+x^4 \right)\\
& 1+x^{4m+2}= (1+x^2)\prod_{k=1}^m \left(1+2x^2\cos\frac{2k\pi}{2m+1}+x^4\right)
\end{align}
and made use of the post that
$$\int_0^\infty \frac{\ln(1+2x^2\cos \theta +x^4)}{1+x^2}dx
=2\pi \ln\left(2\cos\frac{\theta}4\right)
$$
to build up two decent formula for ALL even powers as below:
$$\begin{align}
&\int_0^\infty \frac{\ln(1+x^{4m})}{1+x^2}dx
=2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{(2k-1)\pi}{8m}\bigg)\
\\
&\int_0^\infty \frac{\ln(1+x^{4m+2})}{1+x^2}dx
= \pi\ln2 + 2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{k\pi}{2(2m+1)}\bigg)
\end{align}$$
which can be merged into a single formula below:
$$
\boxed{\int_0^\infty \frac{\ln(1+x^{2n})}{1+x^2}dx =\frac{1-(-1)^n}{2} \pi \ln 2+2 \pi \ln \left[2^{\left[\frac{n}{2}\right]} \prod_{k=1}^{\left[\frac{n}{2}\right]}\left(\cos \frac{(n-2 k+1) \pi}{4 n}\right)\right]}
$$
where $n>1.$
Though the nut for odd powers is much harder, @J.G had cracked the nut partially by giving it a closed form with double sum.
$$\boxed{\int_0^\infty\frac{\ln(1+x^n)}{1+x^2}dx=\frac{\pi}{4}\ln2+G+\sum_{k=0}^{(n-3)/2}\left(\pi\ln\left|2\sin\frac{\pi(2k+1)}{2n}\right|+\frac{\pi(n-4k-2)}{2n}\ln\left|\tan\frac{\pi(2k+1)}{2n}\right|+2\sum_{j\ge0}\frac{(-1)^{j+1}\cos\frac{\pi(2j+1)(2k+1)}{n}}{(2j+1)^2}\right)}$$
Would you like to try to simplify the sum over $j,\,k$?
Backgound
Couple months ago, I met the integrals $$
\int_{0}^{\infty} \frac{\ln \left(x\right)}{1+x^{2}} dx \stackrel{x\mapsto\frac{1}{x}}{=} -\int_{0}^{\infty} \frac{\ln \left(x\right)}{1+x^{2}} d x \Rightarrow \int_{0}^{\infty} \frac{\ln \left(x\right)}{1+x^{2}} dx =0 \tag*{}
$$
$$\textrm{ and}$$
$$
I(0)=\int_{0}^{\infty} \frac{\ln 2}{1+x^{2}} d x=\frac{\pi}{2} \ln 2
$$
Then I started to investigate, in my post 1,
$$I(1)= \int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}dx= \frac{\pi}{4} \ln 2+G $$
Next integral
$$I(2)=\int_{0}^{\infty} \frac{\ln \left(1+x^2\right)}{1+x^{2}} dx \stackrel{x\mapsto\tan {\theta}}{=} -2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta=\pi \ln 2 $$
The third one is split into two integrals, first of which refer to my post 2,
$$\begin{aligned} I(3)&=\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x\\&= \underbrace{\frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}}}_{\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G} d x+\underbrace{\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}}_{\frac{\pi}{4} \ln 2+G} d x\\ &\boxed{\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x =-\frac{G}{3}+\frac{\pi}{4} \ln 2 +\frac{2 \pi}{3}\ln(2+\sqrt{3})}\end{aligned}
$$
The fourth one comes immediately after the my post 3 which states that $$J(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x =\pi\ln \left|2 \cos \frac{a}{2}\right|+|a| \ln \left|\tan \frac{a}{2}\right|-2 \operatorname{sgn} (a) \int_{0}^{\frac{|a|}{2}} \ln (\tan x) d x$$
$$
\begin{aligned}
I(4)&=\int_{0}^{\infty} \frac{\ln \left(1+x^{4}\right)}{1+x^{2}} d x\\&=\int_{0}^{\infty} \frac{\ln \left[\left(x^{2}+\sqrt{2} x+1\right)\left(x^{2}-\sqrt{2} x+1\right)\right]}{1+x^{2}} \\
&=J\left(\frac{\pi}{4}\right)+J\left(-\frac{\pi}{4}\right)\\& =2 \pi \ln \left(2 \cos \frac{\pi}{8}\right)\\& \boxed{\int_{0}^{\infty} \frac{\ln \left(1+x^{4}\right)}{1+x^{2}} d x =\pi \ln (2+\sqrt{2})}
\end{aligned}
$$
The harder one comes from the post 4 by @Quanto.
$$ \begin{aligned}I(5)&=\int_0^1\frac{\ln(1+x^5)}{1+x^2}dx\\
&\boxed{= \frac15G -\frac{19\pi }{20}\ln 2+\frac{3\pi }{5}\ln 5+\frac{4\pi}5 \ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\frac{8\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)}\end{aligned}
$$
When the power of $x$ is raised to the power $6$, I first split the integral into 2.
$$\int_0^1\frac{\ln(1+x^6)}{1+x^2}dx= \underbrace{\int_{0}^{\infty} \frac{\ln \left(x^{2}+1\right)}{1+x^{2}} d x}_{=\pi \ln 2}+ \underbrace{\int_{0}^{\infty} \frac{\ln \left(x^{4}-x^{2}+1\right)}{1+x^{2}} d x
}_{K} $$
For integral $K$, we use the formula founded in my post 3 stating that $$J(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x =\pi\ln \left|2 \cos \frac{a}{2}\right|+|a| \ln \left|\tan \frac{a}{2}\right|-2 \operatorname{sgn} (a) \int_{0}^{\frac{|a|}{2}} \ln (\tan x) d x$$
Then $$
\begin{aligned}
K &=\int_{0}^{\infty} \frac{\ln \left(x^{4}-x^{2}+1\right)}{1+x^{2}} d x \\
&=\int_{0}^{\infty} \frac{\ln \left(x^{2}+\sqrt{3} x+1\right)}{1+x^{2}} d x+\int_{0}^{\infty} \frac{\ln \left(x^{2}-\sqrt{3} x+1 \right) }{1+x^{2}} dx\\
&=J\left(\frac{\pi}{3}\right)+J\left(-\frac{\pi}{3}\right) \\
&=2 \pi \ln \left(2 \cos \frac{\pi}{6}\right) \\
&=\pi \ln 3
\end{aligned}
$$
$$
\boxed{ \int_{0}^{\infty} \frac{\ln \left(1+x^{6}\right)}{1+x^{2}} d x=\pi \ln 2+\pi \ln 3=\pi \ln 6 }
$$
From my recent post, $$
\boxed{\int_{0}^{\infty} \frac{\ln \left(x^{4}+b x^{2}+1\right)}{1+x^{2}} d x = 2 \pi \ln \left[2 \cos \left(\frac{1}{4} \cos ^{-1}\left(\frac{b}{2}\right)\right)\right]}
$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(x^{8}+1\right)}{1+x^{2}}=& \int_{0}^{\infty} \frac{\ln \left(x^{4}+\sqrt{2} x+1\right)}{1+x^{2}} d x +\int_{0}^{\infty} \frac{\ln \left(x^{4}-\sqrt{2} x+1\right)}{1+x^{2}} d x \\
=& 2 \pi \ln \left[2 \cos \frac{1}{4} \cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)\right] +2 \pi \ln \left[2 \cos \frac{1}{4} \cos ^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right] \\
=& 2 \pi \ln \left(2 \cos \frac{\pi}{16}\right)+2 \pi \ln \left(2 \cos \frac{3 \pi}{16}\right)\\
=& 2 \pi \ln \left(2^2 \cos \frac{\pi}{16} \cos \frac{3 \pi}{16}\right)\\=& 2 \pi \ln (\sqrt{2}+\sqrt{2+\sqrt{2}})
\end{aligned}
$$
Through the investigation, I found that it is very interesting to find their exact values as most of them are unexpectedly simple and decent except $I(5)$. We can foresee the difficulty of the evaluation will be increased.
My question is whether we can go further for $n=7$ and $n\geq 9.$ Your help and contributions will be highly appreciated.
Best Answer
For even $n$’s, utilize the factorizations \begin{align} &1+x^{4m}= \prod_{k=1}^m \left(1+2x^2\cos\frac{(2k-1)\pi}{2m}+x^4 \right)\\ & 1+x^{4m+2}= (1+x^2)\prod_{k=1}^m \left(1+2x^2\cos\frac{2k\pi}{2m+1}+x^4\right) \end{align} and the result $$\int_0^\infty \frac{\ln(1+2x^2\cos \theta +x^4)}{1+x^2}dx =2\pi \ln\left(2\cos\frac{\theta}4\right) $$ to obtain \begin{align} &\int_0^\infty \frac{\ln(1+x^{4m})}{1+x^2}dx =2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{(2k-1)\pi}{8m}\bigg)\ \\ &\int_0^\infty \frac{\ln(1+x^{4m+2})}{1+x^2}dx = \pi\ln2 + 2\pi \ln \bigg( 2^m \prod_{k=1}^m \cos\frac{k\pi}{2(2m+1)}\bigg) \end{align} In particular \begin{align} \int_0^\infty \frac{\ln(1+x^{4})}{1+x^2}dx =& \ \pi\ln(2+\sqrt2)\\ \int_0^\infty \frac{\ln(1+x^{6})}{1+x^2}dx =& \ \pi\ln6\\ \int_0^\infty \frac{\ln(1+x^{8})}{1+x^2}dx =& \ 2\pi\ln\left(\sqrt2+\sqrt{2+\sqrt2}\right)\\ \int_0^\infty \frac{\ln(1+x^{10})}{1+x^2}dx =&\ \pi\ln(10+4\sqrt5)\\ \int_0^\infty \frac{\ln(1+x^{12})}{1+x^2}dx =& \ \pi\ln(2+\sqrt2)+2\pi \ln(\sqrt3+\sqrt2)\\ \int_0^\infty \frac{\ln(1+x^{14})}{1+x^2}dx =&\ \pi\ln(14) + 2\pi \ln\cot\frac\pi{7}\\ \int_0^\infty \frac{\ln(1+x^{16})}{1+x^2}dx =&\ 2\pi\ln\bigg( 2+ \sqrt2 +\sqrt{2+\sqrt2}\\ &\hspace{10mm} + \sqrt{2(2+\sqrt2)\left(2+ \sqrt{2+\sqrt2}\right)}\bigg)\\ \int_0^\infty \frac{\ln(1+x^{18})}{1+x^2}dx =&\ \pi\ln 6+2\pi\ln\cot\frac\pi{18}\\ \int_0^\infty \frac{\ln(1+x^{20})}{1+x^2}dx =&\ 2\pi\bigg( \ln\sqrt{2+\sqrt2} + \ln\bigg(1 +\frac{\sqrt5+1}{2\sqrt2}\bigg)\\ &\hspace{10mm} + \ln\left(2 +\sqrt{5+\sqrt5}\right) \bigg)\\ \end{align}